Question

19. An aircraft maintaining a constant airspeed of 525 MPH is headed due south. The jet stream is

120 MPH in the northeasterly direction.

a. Sketch the scenario on the x-y axis.

b. Express the velocity of the plane relative to the air as a vector.

c. Express the velocity of the jet stream as a vector.

d. Find the velocity of the plane relative to the ground.

Find the actual speed and direction of the plane relative to the ground. Round to the

tenths.

e.

Answer 1

a. Sketch is given in the attachment

b.
`V_(p)` = (0,-525)

c.
`V_(w)` = (
`√(2)`,
`√(2)`)

d.
`V_(g)` = (60
`√(2)`, 60
`√(2)` - 525)

e. Actual Speed = |
`V_(g)`| = 448.3 MPH

Direction = Ф = - 79.1°

Direction = Ф = 280.9° with respect to x - axis.

Explanation:

a. Sketch the scenario on the x-y axis.

Where,

Velocity of the plane relative to air =
`V_(p)`

Velocity of the jet stream =
`V_(w)`

Velocity of the plane relative to ground =
`V_(g)`

Solution:

Note: Solution is given in the attachment. Please refer to the attachment for the sketch.

b. Express the velocity of the plane relative to the air as a vector.

Solution:

`V_(p)` = |
`V_(p)`| cosФi + |
`V_(p)`| sinФj

where,

|
`V_(p)`| = magnitude of the vector.

|
`V_(p)`| = 525 MPH

Ф = 270° w.r.t to x - axis. (See the sketch in the attachment)

`V_(p)` = |
`V_(p)`| cosФi + |
`V_(p)`| sinФj

Plug in the values into this equation to express it in the vector form as required.

`V_(p)` = (525) cos270i + (525) sin270j

As, Cos270 = 0

and

Sin270 = -1

So,

`V_(p)` = (525) (0)i + (525) (-1)j

`V_(p)` = 0i -525j

`V_(p)` = (0,-525)

c. Express the velocity of the jet stream as a vector.

Solution:

Velocity of the jet stream as a vector.

`V_(w)` = |
`V_(w)`| cosФi + |
`V_(w)`| sinФj

where,

|
`V_(w)`| = magnitude of the vector.

|
`V_(p)`| = 120 MPH

Ф = 45° (See the sketch in the attachment)

`V_(w)` = |
`V_(w)`| cosФi + |
`V_(w)`| sinФj

Plug in the values into this equation to express it in the vector form as required.

`V_(w)` = (120) cos45i + (120) sin45j

As, Cos45 =
`(1)/(√(2) )`

and

Sin45 =
`(1)/(√(2) )`

So,

`V_(w)` = (120) (
`(1)/(√(2) )`)i + (120) (
`(1)/(√(2) )`)j

`V_(w)` = 60
`√(2)`i +60
`√(2)`j

`V_(w)` = (
`√(2)`,
`√(2)`)

d. Find the velocity of the plane relative to the ground.

Solution:

Velocity of the plane relative to the ground = Sum of velocity of plane relative to the air and velocity of the jet stream

`V_(g)` =
`V_(p)` +
`V_(w)`

`V_(g)` = 0i -525j + 60
`√(2)`i +60
`√(2)`j

`V_(g)` = 60
`√(2)`i + 60
`√(2)` - 525j

`V_(g)` = (60
`√(2)`, 60
`√(2)` - 525)

e. Find the actual speed and direction of the plane relative to the ground. Round to the tenths.

Solution:

Actual Speed = |
`V_(g)`| = magnitude of the velocity of the plane relative to the ground.

Actual Speed = |
`V_(g)`| =
`\sqrt{(60√(2)) ^(2) + (60√(2)-525) ^(2) }`

Actual Speed = |
`V_(g)`| = 448.3 MPH

Direction = Ф =
`tan^(-1)` (
`(60√(2)-525 )/(60√(2) )`)

Direction = Ф = - 79.1°

To find out the direction with respect to x - axis.

Direction = Ф = 360 - 79.1

Direction = Ф = 280.9°